There is a concept known as eective area which is useful when dealing with antennas in the receiving mode.If an antenna is irradiated by an electromagnetic wave, the received power avai
Trang 1So, given PTand G it is possible to calculate the power density at any point in the far ®eld that lies in the direction of maximum radiation.A knowledge of the radiation pattern is necessary to determine the power density at other points The power gain is unity for an isotropic antenna, i.e one which radiates uniformly
in all directions, and an alternative de®nition of power gain is therefore the ratio of power density, from the speci®ed antenna, at a given distance in the direction of maximum radiation, to the power density at the same point, from an isotropic antenna which radiates the same power.As an example, the power gain of a half-wave dipole is 1.64 (2.15 dB) in a direction normal to the dipole and is the same whether the antenna is used for transmission or reception
There is a concept known as eective area which is useful when dealing with antennas in the receiving mode.If an antenna is irradiated by an electromagnetic wave, the received power available at its terminals is the power per unit area carried
by the wave6the eective area, i.e P WA.It can be shown [1, Ch.11] that the eective area of an antenna and its power gain are related by
A l2G
2.2 PROPAGATION IN FREE SPACE
Radio propagation is a subject where deterministic analysis can only be applied in a few rather simple cases.The extent to which these cases represent practical conditions is a matter for individual interpretation, but they do give an insight into the basic propagation mechanisms and establish bounds
If a transmitting antenna is located in free space, i.e remote from the Earth or any obstructions, then if it has a gain GT in the direction to a receiving antenna, the power density (i.e power per unit area) at a distance (range) d in the chosen direction is
The available power at the receiving antenna, which has an eective area A is therefore
PRP4pdTG2T A
P4pdTG2T l24pGR
where GRis the gain of the receiving antenna
Thus, we obtain
PR
PT GTGR
l 4pd
2
2:3
Trang 2which is a fundamental relationship known as the free space or Friis equation [2].The well-known relationship between wavelength l, frequency f and velocity of propagation c (c f l) can be used to write this equation in the alternative form
PR
PT GTGR
c 4pfd
2:4
The propagation loss (or path loss) is conveniently expressed as a positive quantity and from eqn.(2.4) we can write
LF dB 10 log10 PT=PR
10 log10GT 10 log10GR 20 log10 f 20 log10d k 2:5
where k 20 log10 4p
3 108
147:56
It is often useful to compare path loss with the basic path loss LBbetween isotropic antennas, which is
LB dB 32:44 20 log10 fMHz 20 log10dkm 2:6
If the receiving antenna is connected to a matched receiver, then the available signal power at the receiver input is PR.It is well known that the available noise power is kTB, so the input signal-to-noise ratio is
SNRi PR
kTB
PTGTGR kTB
c 4p fd
If the noise ®gure of the matched receiver is F, then the output signal-to-noise ratio is given by
SNRo SNRi=F
or, more usefully,
SNRodB SNRidB FdB Equation (2.4) shows that free space propagation obeys an inverse square law with range d, so the received power falls by 6 dB when the range is doubled (or reduces by
20 dB per decade).Similarly, the path loss increases with the square of the transmission frequency, so losses also increase by 6 dB if the frequency is doubled High-gain antennas can be used to make up for this loss, and fortunately they are relatively easily designed at frequencies in and above the VHF band.This provides a solution for ®xed (point-to-point) links, but not for VHF and UHF mobile links where omnidirectional coverage is required
Sometimes it is convenient to write an expression for the electric ®eld strength at a known distance from a transmitting antenna rather than the power density.This can
be done by noting that the relationship between ®eld strength and power density is
Trang 3W EZ2 where Z is the characteristic wave impedance of free space.Its value is 120p (377 O) and so eqn.(2.2) can be written
E2 120p
PTGT 4pd2 giving
E
30PTGT p
Finally, we note that the maximum useful power that can be delivered to the terminals of a matched receiver is
P E2A
E2 120p
l2GR 4p
El 2p
2
GR
2.3 PROPAGATION OVER A REFLECTING SURFACE
The free space propagation equation applies only under very restricted conditions; in practical situations there are almost always obstructions in or near the propagation path or surfaces from which the radio waves can be re¯ected.A very simple case, but one of practical interest, is the propagation between two elevated antennas within line-of-sight of each other, above the surface of the Earth.We will consider two cases, ®rstly propagation over a spherical re¯ecting surface and secondly when the distance between the antennas is small enough for us to neglect curvature and assume the re¯ecting surface to be ¯at.In these cases, illustrated in Figures 2.1 and 2.4 the received signal is a combination of direct and ground-re¯ected waves To determine the resultant, we need to know the re¯ection coecient
2.3.1 The re¯ection coecient of the Earth
The amplitude and phase of the ground-re¯ected wave depends on the re¯ection coecient of the Earth at the point of re¯ection and diers for horizontal and vertical polarisation.In practice the Earth is neither a perfect conductor nor a perfect dielectric, so the re¯ection coecient depends on the ground constants, in particular the dielectric constant e and the conductivity s
For a horizontally polarised wave incident on the surface of the Earth (assumed to
be perfectly smooth), the re¯ection coecient is given by [1, Ch.16]:
rhsin c
e=e0 js=oe0 cos2c p
sin c p e=e0 js=oe0 cos2c where o is the angular frequency of the transmission and e0is the dielectric constant
of free space.Writing eras the relative dielectric constant of the Earth yields
Trang 4rhsin c
er jx cos2c p
sin c p er jx cos2c 2:9 where
x oes
018 10f 9s For vertical polarisation the corresponding expression is
rv er j x sin c p er jx cos2c
er jx sin c p er jx cos2c 2:10 The re¯ection coecients rhand rvare complex, so the re¯ected wave will dier from the incident wave in both magnitude and phase.Examination of eqns (2.9) and (2.10) reveals some quite interesting dierences For horizontal polarisation the relative phase of the incident and re¯ected waves is nearly 1808 for all angles of incidence.For very small values of c (near-grazing incidence), eqn.(2.9) shows that the re¯ected wave is equal in magnitude and 1808 out of phase with the incident wave for all frequencies and all ground conductivities.In other words, for grazing incidence
rh jrhjejy 1ej p 1 2:11
As the angle of incidence is increased then jrhj and y change, but only by relatively small amounts.The change is greatest at higher frequencies and when the ground conductivity is poor
Figure 2.1 Two mutually visible antennas located above a smooth, spherical Earth of eective radius re
Trang 5For vertical polarisation the results are quite dierent.At grazing incidence there
is no dierence between horizontal and vertical polarisation and eqn.(2.11) still applies.As c is increased, however, substantial dierences appear.The magnitude and relative phase of the re¯ected wave decrease rapidly as c increases, and at an angle known as the pseudo-Brewster angle the magnitude becomes a minimum and the phase reaches 7908.At values of c greater than the Brewster angle, jrvj increases again and the phase tends towards zero.The very sharp changes that occur
in these circumstances are illustrated by Figure 2.2, which shows the values of jrvj and y as functions of the angle of incidence c.The pseudo-Brewster angle is about
158 at frequencies of interest for mobile communications (x er), although at lower frequencies and higher conductivities it becomes smaller, approaching zero if x er Table 2.1 shows typical values for the ground constants that aect the value of r The conductivity of ¯at, good ground is much higher than the conductivity of poorer ground found in mountainous areas, whereas the dielectric constant, typically 15, can be as low as 4 or as high as 30.Over lakes or seas the re¯ection properties are quite dierent because of the high values of s and er.Equation (2.11) applies for horizontal polarisation, particularly over sea water, but r may be signi®cantly dierent from 71 for vertical polarisation
Figure 2.2 Magnitude and phase of the plane wave re¯ection coecient for vertical polarisation.Curves drawn for s 12 10 3, er 15.Approximate results for other frequencies and conductivities can be obtained by calculating the value of x as 18 103s=fMHz
Trang 62.3.2 Propagation over a curved re¯ecting surface
The situation of two mutually visible antennas sited on a smooth Earth of eective radius re is shown in Figure 2.1 The heights of the antennas above the Earth's surface are hTand hR, and above the tangent plane through the point of re¯ection the heights are h0T and h0R.Simple geometry gives
d2 re hT h0
T2 r2 hT h0
T2 2re hT h0
T ' 2re hT h0
T 2:12 and similarly
d2' 2re hR h0
Using eqns (2.12) and (2.13) we obtain
h0
T hT d21
2re and h0R hR
d2 2
The re¯ecting point, where the two angles marked c are equal, can be determined by noting that, providing d1, d244hT, hR, the angle c (radians) is given by
c hdT0
1 hd0R 2 Hence
h0 T
h0R'dd1
Using the obvious relationship d d1+d2 together with equations (2.14) and (2.15) allows us to formulate a cubic equation in d1:
2d3
1 3dd2 d2 2re hT hRd1 2rehTd 0 2:16 The appropriate root of this equation can be found by standard methods starting from the rough approximation
d1'1 hd
T=hR
To calculate the ®eld strength at a receiving point, it is normally assumed that the dierence in path length between the direct wave and the ground-re¯ected wave is negligible in so far as it aects attenuation, but it cannot be neglected with regard to the phase dierence along the two paths.The length of the direct path is
Table 2.1 Typical values of ground constants
Surface Conductivity s (S) Dielectric constant er Poor ground (dry) 11073 4±7
Good ground (wet) 21072 25±30
Trang 7R1 d 1 hT0 d2h0R2
and the length of the re¯ected path is
R2 d 1 hT0 hd2 0R2
The dierence DR R2 R1 is
DR d 1 hT0 h0
R2
d2
1 h0T h0
R2
d2
and if d hT0, hR0 this reduces to
The corresponding phase dierence is
Df 2pl DR 4phld0Th0R 2:18
If the ®eld strength at the receiving antenna due to the direct wave is Ed, then the total received ®eld E is
E Ed1 r exp j Df
where r is the re¯ection coecient of the Earth and r j rjexp jy.Thus,
E Edf1 jrjexp j Df yg 2:19 This equation can be used to calculate the received ®eld strength at any location, but note that the curvature of the spherical Earth produces a certain amount of divergence of the ground-re¯ected wave as Figure 2.3 shows This eect can be taken into account by using, in eqn.(2.19), a value of r which is dierent from that derived
in Section 2.3.1 for re¯ection from a plane surface The appropriate modi®cation consists of multiplying the value of r for a plane surface by a divergence factor D given by [3]:
D ' 1 r 2d1d2
e h0T h0R
2:20
The value of D can be of the order of 0.5, so the eect of the ground-re¯ected wave is considerably reduced
2.3.3 Propagation over a plane re¯ecting surface
For distances less than a few tens of kilometres, it is often permissible to neglect Earth curvature and assume the surface to be smooth and ¯at as shown in Figure 2.4 If we also assume grazing incidence so that r 1, then eqn.(2.19) becomes
Trang 8E Ed1 exp jDf
Ed1 cos Df j sin Df
Thus,
jEj jEdj1 cos2Df 2 cos Df sin2Df1=2
2jEdjsinDf2 and using eqn.(2.18), with hT0 hTand h0R hR,
jEj 2jEdjsin 2phThR
ld
The received power PRis proportional to E2so
PR/ 4jEdj2sin2 2phThR
ld
4PT 4pdl
2
GTGRsin2 2phThR
ld
2:21
If d44hT, hRthis becomes
PR
PT GTGR hTdh2R
2:22 Figure 2.3 Divergence of re¯ected rays from a spherical surface
Figure 2.4 Propagation over a plane earth
Trang 9Equation (2.22) is known as the plane earth propagation equation It diers from the free space relationship (2.3) in two important ways First, since we assumed that d44hT, hR, the angle Df is small and l cancels out of eqn.(2.22), leaving it frequency independent.Secondly, it shows an inverse fourth-power law with range rather than the inverse square law of eqn.(2.3).This means a far more rapid decrease
in received power with range, 12 dB for each doubling of distance in this case Note that eqn.(2.22) only applies at ranges where the assumption d44hT, hR is valid.Close to the transmitter, eqn.(2.21) must be used and this gives alternate maxima and minima in the signal strength as shown in Figure 2.5
In convenient logarithmic form, eqn.(2.22) can be written
LP dB 10 log10 PT=PR
10 log10GT 10 log10GR 20 log10hT 20 log10hR 40 log10d
2:23 and by comparison with eqn (2.6) we can write a `basic loss' LBbetween isotropic antennas as
LB dB 40 log10d 20 log10hT 20 log10hR 2:24
2.4 GROUND ROUGHNESS
The previous section presupposed a smooth re¯ecting surface and the analysis was therefore based on the assumption that a specular re¯ection takes place at the point where the transmitted wave is incident on the Earth's surface.When the surface is Figure 2.5 Variation of signal strength with distance in the presence of specular re¯ection
Trang 10rough the specular re¯ection assumption is no longer realistic since a rough surface presents many facets to the incident wave.A diuse re¯ection therefore occurs and the mechanism is more akin to scattering.In these conditions characterisation by a single complex re¯ection coecient is not appropriate since the random nature of the surface results in an unpredictable situation.Only a small fraction of the incident energy may be scattered in the direction of the receiving antenna, and the `ground-re¯ected' wave may therefore make a negligible contribution to the received signal
In these circumstances it is necessary to de®ne what constitutes a rough surface Clearly a surface that might be considered rough at some frequencies and angles of incidence may approach a smooth surface if these parameters are changed.A measure of roughness is needed to quantify the problem, and the criterion normally used is known as the Rayleigh criterion.The problem is illustrated in Figure 2.6(a) and an idealised rough surface pro®le is shown in Figure 2.6(b)
Consider the two rays A and B in Figure 2.6(b) Ray A is re¯ected from the upper part of the rough surface and ray B from the lower part.Relative to the wavefront
AA0shown, the dierence in path length of the two rays when they reach the points
C and C0after re¯ection is
Dl AB BC A0B0 B0C0
sin cd 1 cos 2c
Figure 2.6 Re¯ections from a semi-rough surface: (a) practical terrain situation, (b) idealised model